How to Prove Something is a Contradiction only by Logical Equivalencies Having the proposition:
$$
(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\land(p \rightarrow q) \land (p \rightarrow \lnot q)
$$
I want to prove this to be contradiction. So far I have that:
$$
p \rightarrow q \equiv \lnot p \lor q
$$
But now I am stuck because I want to proceed with the Distributive Law, but don't know how to apply it in my new situation:
$$
(p \lor q) \land (p \lor \lnot q) \land (\lnot p \lor q) \land (\lnot p \lor \lnot q)
$$
 A: By the distributive rule, we have:
$$\begin{align}(p \lor q) \land (p \lor \lnot q) \land (\lnot p \lor q) \land (\lnot p \lor \lnot q)&\equiv [p \lor (q\land \lnot q)] \land [\lnot p \lor (q \land \lnot q)]\\ \\
&\equiv (p \lor \bot) \land (\lnot p \lor \bot)\\ \\
&\equiv (p \land \lnot p)\\ \\
&\equiv \bot\end{align}$$
Note that $\bot$ means "logically false," aka a "contradiction".
A: We are going to prove that
$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\equiv p\;.$
Indeed,
$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\equiv (p\lor q)\land(p\lor\lnot q)\equiv\\\equiv p\lor(q\land\lnot q)\equiv p\lor\text{False}\equiv p\;.$
Moreover it results that
$(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv \lnot p\;.$
Indeed,
$(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv (\lnot p\lor q)\land(\lnot p\lor\lnot q)\equiv\\\equiv \lnot p\lor(q\land\lnot q)\equiv \lnot p\lor\text{False}\equiv \lnot p\;.$
Hence,
$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\land(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv\\\equiv p\land\lnot p\equiv\text{False}\;.$
